FCC Exam Question: 6A79

How many capacitors of 400 volts and 2 microfarad each would be necessary to obtain a combinationrated at 1600 volts and 1.5 microfarad?

A. 10
B. 12
C. 14
D. 16
Correct Answer: B

Explanation: To achieve a higher voltage rating, capacitors must be connected in series. The total voltage rating for capacitors in series is the sum of the individual voltage ratings. Desired voltage = 1600 V Individual capacitor voltage = 400 V Number of capacitors in series (N_s) = 1600 V / 400 V = 4 capacitors. When capacitors are connected in series, their total capacitance decreases. For N_s equal capacitors in series, the total capacitance of that string is C_string = C_individual / N_s. C_string = 2 microfarad / 4 = 0.5 microfarad. We need a total capacitance of 1.5 microfarad. To increase capacitance, strings of capacitors must be connected in parallel. Number of parallel strings (N_p) = Desired total capacitance / Capacitance of one string N_p = 1.5 microfarad / 0.5 microfarad = 3 parallel strings. Therefore, we need 3 parallel strings, with each string containing 4 capacitors in series. Total number of capacitors = N_s * N_p = 4 capacitors/string * 3 strings = 12 capacitors.

Pass Your FCC Exam!

Study offline, track your progress, and simulate real exams with the GMDSS Trainer app.


Includes Elements 1, 3, 6, 7R, 8, and 9.